If OT and ON are perpendiculars dropped from the origin to the tangent and normal to the curve $x=a \sin ^3 t, y=a \cos ^3 t$ at an arbitrary point, then which one of the following is correct ?
Answer & explanation
Correct answer: option 4
At any point 't' on the given curve, we have
$\frac{d y}{d x}=\frac{d y}{d t} / \frac{d x}{d t}=\frac{3 a \cos ^2 t(-\sin t)}{3 a \sin ^2 t(\cos t)}=-\cot t$
The equations of the tangent and normal at 't' are
$x \cos t+y \sin t-\frac{a}{2} \sin 2 t=0$ .......(i)
and, $x \sin t-y \cos t+a \cos 2 t=0$ .......(ii)
∴ OT = $\frac{|(a / 2) \sin 2 t|}{\sqrt{\cos ^2 t+\sin ^2 t}}=\frac{a}{2} \sin 2 t \Rightarrow 2 O T=a \sin 2 t$
and, ON = $\frac{|a \cos 2 t|}{\sqrt{\sin ^2 t+\cos ^2 t}}=a \cos 2 t$
Hence,
$4 O T^2+O N^2=a^2 \sin ^2 2 t+a^2 \cos ^2 2 t=a^2$
Length of the tangent at 't' = $\frac{\left|y \sqrt{1+\left(\frac{d y}{d x}\right)^2}\right|}{d y / d x}$
⇒ Length of the tangent at 't' = $\left|\frac{y ~cosec~t}{-\cot t}\right|=\left(\frac{y}{\cos t}\right)$
Length of the normal at 't' $=\left|y \sqrt{1+\left(\frac{d y}{d x}\right)^2}\right|$
⇒ Length of the normal at 't' = $\left|y \sqrt{1+\cot ^2 t}\right|=\left|\frac{y}{\sin t}\right|$
Hence, options (a), (b) and (c) are correct.